3 ways to find perfect numbers in a range in Java

How to find perfect numbers in a range in Java:

This post will explain how to write a Java program to find all perfect numbers in a given range. You will understand how to check if a number is a perfect number or not and how to find all perfect numbers in a given range.

What is a perfect number:

A number is called a perfect number if the number is equal to the sum of its factors, excluding the number itself.

For example, 6 is a perfect number because the sum of its factors is 1 + 2 + 3 which is equal to 6.

So, we need to find the factors of a number and by comparing the sum of the factors with the number itself, we can say if the number is a perfect number or not.

Example 1: Java program to find the perfect numbers in a given range:

The below program shows how to find all the perfect numbers in a user-given range:

import java.util.Scanner;

public class Main {

    private static boolean isPerfect(int n) {
        int sum = 1;
        for (int i = 2; i < n; i++) {
            if (n % i == 0)
                sum += i;
        }
        return sum == n && n != 1;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int start, end;

        System.out.println("Enter the start of the range: ");
        start = scanner.nextInt();

        System.out.println("Enter the end of the range: ");
        end = scanner.nextInt();

        System.out.println("Perfect numbers in between " + start + " and " + end + " are:");
        for (int no = start; no <= end; no++) {
            if (isPerfect(no)) {
                System.out.println(no);
            }
        }
    }
}

It uses a Scanner variable to read the start and end of the range.
It uses a for loop to find all perfect numbers between start and end(inclusive).
On each iteration, it calls the isPerfect method to check if the number is a perfect number or not.
The isPerfect method takes one number as its parameter and returns one boolean value defining if the number is a perfect number or not.
- It initializes one variable sum as 1 to hold the sum of the factors of the number. It is initialized as 1 as 1 is the factor of all numbers.
- It runs one for loop from i = 2 to i = n - 1, where n is the given number.
- On each iteration, it checks if i can divide the number n or not. If yes, it will add that value to the sum variable.
At the end of this method, it returns true if the calculated sum is equal to the given number and if it is not equal to 1. Else, it returns false.

Example 2: Java program to find the perfect numbers in a given range by iterating up to half of the number:

We can reduce the iteration by half. Instead of iterating from 2 to number - 1, we can iterate from 2 to number/2 as larger than this value can’t be a factor of the number.

The final program become:

import java.util.Scanner;

public class Main {

    private static boolean isPerfect(int n) {
        int sum = 1;
        for (int i = 2; i <= n/2; i++) {
            if (n % i == 0)
                sum += i;
        }
        return sum == n && n != 1;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int start, end;

        System.out.println("Enter the start of the range: ");
        start = scanner.nextInt();

        System.out.println("Enter the end of the range: ");
        end = scanner.nextInt();

        System.out.println("Perfect numbers in between " + start + " and " + end + " are:");
        for (int no = start; no <= end; no++) {
            if (isPerfect(no)) {
                System.out.println(no);
            }
        }
    }
}

The only change in this program is:

...
...
for (int i = 2; i <= n/2; i++) {
    ...

It will work similarly to the above program but it will be faster.

Example 3: Recursive Java program to find the perfect numbers in a given range:

Let’s write down the program recursively. The isPerfect method will recursively check if a number is perfect or not.

import java.util.Scanner;

public class Main {

    private static boolean isPerfect(int n, int i, int sum) {
        if(n == 1)
            return false;

        if (i > n / 2)
            return sum == n;

        if (n % i == 0)
            return isPerfect(n, i + 1, sum + i);
        else
            return isPerfect(n, i + 1, sum);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int start, end;

        System.out.println("Enter the start of the range: ");
        start = scanner.nextInt();

        System.out.println("Enter the end of the range: ");
        end = scanner.nextInt();

        System.out.println("Perfect numbers in between " + start + " and " + end + " are:");
        for (int no = start; no <= end; no++) {
            if (isPerfect(no, 2, 1)) {
                System.out.println(no);
            }
        }
    }
}

The isPerfect method is changed to take three parameters. It takes n, i and sum variables as the parameters.
This method is called with i = 2 and sum = 1 in the main method.
Once the value of i increases more than number/2, it checks if the value of sum and n are equal or not and returns that value. Else, it checks if the current value of i is a factor of n or not and calls the isPerfect method recursively.

It will give similar output.

Enter the start of the range: 
1
Enter the end of the range: 
10000
Perfect numbers in between 1 and 10000 are:
6
28
496
8128

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